Answer to Question #172678 in Calculus for Todd Phillips

Question

For the function f(x) = 4x³ – x²– 3, determine

a. The intervals of increase or decrease

b. The location of any maximum or minimum points c. The intervals of concavity up or down

d. The location of any points of inflection

Answer 1

The first derivative of the function is f'(x) = 12x²-2x = 2x(6x-1).

a. If the first derivative f'(x)>0 function is increase. Solving of 2x(6x-1)>0 is (-∞,0) and (1/6,+∞). That's why function increase at the intervals : (-∞,0) and (1/6,+∞).

In the interval (0,1/6) function decrease.

b. Maximum and minimum points is points where f'(x) = 0. The solving of equation 2x(6x-1) is: x₁ = 0 and x₂ =1/6. f(0) = -3; f(1/6) = -325/108.

The intervals of concavity up or down depend on the sign of the second derivative.

the second derivative is f''(x) = 24x-2 = 2(12x-1) . f''>0 at the interval (1/12,+∞). Thats why f(x) is concavity up at the interval (1/12,+∞) and concavity down at (-∞,1/12).

c. The location of any points of inflection depends on the second derivative. f''=0.

Solving of 2(12x-1) = 0 is x=1/12. f(x) = -649/216. This is point of inflection.